Intuition: Why is continuous decay expressed as the inverse of the equivalent continuous growth rate?

Question

I understand $e$ as $\lim_{n \to \infty} \big(1+\frac{1}{n}\big)^n$. I also (finally) understand the idea that continuous growth is "a rate that is applied constantly to the amount present at any instant" (see Confusion about Continuous Growth being "a rate that is applied constantly to the amount present at any instant").

I'm trying to understand continuous decay in terms of / compared to (and maybe this is my problem) continuous growth. I think I have decent physical intuition as to what it "means", however what is not obvious to me is why it is expressed as $\frac{1}{e}$. It vaguely seems to make sense (division as opposed to multiplication...), but it's not clear at all.

I've also derived it algebraically as such:

Let $t = -n$, then $$\bigg(1-\frac{1}{n}\bigg)^n = \bigg(1 + \frac{1}{t}\bigg)^{-t} = \Bigg[\bigg(1 + \frac{1}{t}\bigg)^{t}\Bigg]^{-1}$$ so that

$$\lim_{t \to \infty} \Bigg[\bigg(1 + \frac{1}{t}\bigg)^{t}\Bigg]^{-1} = e^{-1} = \frac{1}{e}$$

However, I find little comfort and intuition in the algebra.

Is there a way to think about this that makes it obvious why continuous decay is expressed as the inverse of the equivalent continuous growth rate?

Or in other words, why $\lim_{n \to \infty}\big(1-\frac{1}{n}\big)^n = \frac{1}{e}$

Answer 1

Ok, so this is what I've come up with.

You can express $1+\frac{1}{x}$ as $\frac{x+1}{x}$, such that $\big(1+\frac{1}{x}\big)^x = \big(\frac{x+1}{x}\big)^x$

And you can express $1-\frac{1}{x}$ as $\frac{x-1}{x}$, such that $\big(1-\frac{1}{x}\big)^x = \big(\frac{x-1}{x}\big)^x =\frac{1}{\big(\frac{x}{x-1}\big)^x}$

Now $\big(\frac{x}{x-1}\big)^x$ and $\big(\frac{x+1}{x}\big)^x$ look very similar and you can guess (and see on Mathematica) that $\lim_{x \to \infty} \big(\frac{x}{x-1}\big)^x$ also approaches $e$.

Then,

$$\lim_{x \to \infty} \big(\frac{x}{x-1}\big)^x = \lim_{x \to \infty} \big(\frac{x}{x-1}\big)^{x-1} \cdot \lim_{x \to \infty} \big(\frac{x}{x-1}\big)$$ $$= \lim_{x \to \infty} \big(\frac{x}{x-1}\big)^{x-1} \cdot 1$$ $$=\lim_{x \to \infty} \big(\frac{x+1}{x}\big)^x = e$$

Also, taking actual numbers. At $x=10$ you get $$\bigg(1-\frac{1}{10}\bigg)^{10} = \bigg(\frac{9}{10}\bigg)^{10} = \frac{1}{\big(\frac{10}{9}\big)^9 \cdot \frac{10}{9}} = \frac{1}{\big(1+\frac{1}{9}\big)^9 \cdot \frac{10}{9}}$$

Where you can see that as $x$ grows to infinity, the left factor in the denominator converges towards $e$ as the right factor converges to 1.

Answer 2

If the notion of matter can neither be created nor destroyed is true, but merely changed in state, then if you decay matter over time, it seems this could only occur due to the radiation of energy, thus decaying the element to a lesser element. This may be the reason 1/e is necessary. I could be wrong though.

markc